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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Hielly on February 08, 2009, 05:58:19 pm

Title: yr 11 questions
Post by: Hielly on February 08, 2009, 05:58:19 pm: hey :)
Need help on 2 questions

1)Find x.
$ax+by=a^2+2ab-b^2$
$bx+ay=a^2+b^2$

2)Find x and y
3(x-a)-2(y+a)=5-4a
2(x+a)+3(y-a)=4a-1

Thanks
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 08, 2009, 06:21:36 pm: $ax+by=a^2+2ab-b^2$

$bx+ay=a^2+b^2$

times the first equation by a and the second by b

$a^2x+aby=a^3+2a^2b-ab^2$

$b^2x+aby=a^2b+b^3$

now minus the first from the 2nd

$a^2x-b^2x= a^3-a^2b+2a^2b-b^3-ab^2$

$x(a^2-b^2) = a^3+a^2b-b^3-ab^2$

factorise it

$x = \frac{ (a+b)^2(a-b)}{(a-b)(a+b)}$

$x = (a+b)$
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 08, 2009, 06:26:13 pm: $3(x-a)-2(y+a)=5-4a$

$2(x+a)+3(y-a)=4a-1$

expand

$3x-3a-2y-2a=5-4a$

$2x+2a+3y-3a=4a-1$

times the first by 3 and the second by 2

$9x-9a-6y-6a=15-12a$

$4x+4a+6y-6a=8a-2$

add them

$13x-5a-12a=13-4a$

$13x-17a=13-4a$

$13x=13+13a$

$x=1+a$
Title: Re: quick 2 questions on simulanteous equations
Post by: Hielly on February 08, 2009, 06:26:30 pm: Hey flaming_arrow, the answer is x=a+b, there isn't a square root.

1st question
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 08, 2009, 06:29:23 pm: oh whoops typo
fixed.
Title: Re: quick 2 questions on simulanteous equations
Post by: Hielly on February 08, 2009, 06:45:47 pm: $a^3+a^2b-b^3-ab^2$

when factorized i got this, in my previous attempt

$\frac{a^2(a+b)-b^2(b+a)}{(a-b)(a+b)}$

$x = \frac{ (a-b)^2(a+b)}{(a-b)(a+b)}$

$x = (a-b)$

i believe i screwed up there, wondering why this is wrong?
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 08, 2009, 06:51:59 pm: $a^3+a^2b-b^3-ab^2$

$\frac{a^2(a+b)-b^2(b+a)}{(a-b)(a+b)}$

$\frac{ (a+b)(a-b)(a+b)}{(a-b)(a+b)}$

$\implies (a+b)$
Title: Re: quick 2 questions on simulanteous equations
Post by: Hielly on February 08, 2009, 06:53:34 pm: ohhhh silly me, i forgot about (a-b)^2 which equals to (a-b)(a+b)

thanks!!
Title: Re: quick 2 questions on simulanteous equations
Post by: Hielly on February 08, 2009, 07:21:14 pm: okay im trying to solve for y first...

$3(x-a)-2(y+a)=5-4a$

$2(x+a)+3(y-a)=4a-1$

expand

$3x-3a-2y-2a=5-4a$

$2x+2a+3y-3a=4a-1$

times the first by 2 and the second by 3....

$6x-6a-4y-4a=10-8a$

$6x+6a+9y-9a=12a-3$

add them

$-12a-13y+5a=10-12a-8a-3$

$-13y=7-20a-5a+12a$

$-13y=7-13a$

$y=\frac {7-13a}{-13}$

hrm, what did i do wrong here?
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 08, 2009, 07:24:44 pm: i think you mean minus them

anyway

10--3 = 13
Title: Re: quick 2 questions on simulanteous equations
Post by: Hielly on February 08, 2009, 07:27:16 pm: oh so thats how it is, i thought it was like 10-12a and then the 12a takes the minus, so then its 10+ something
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 08, 2009, 07:30:26 pm: you can do it like this

$(10-8a)-(12a-3) = 13-20a$
Title: Re: quick 2 questions on simulanteous equations
Post by: Mao on February 08, 2009, 10:45:46 pm: Quote from: Hielly on February 08, 2009, 06:53:34 pm
ohhhh silly me, i forgot about (a-b)^2 which equals to (a-b)(a+b)

thanks!!

no

$a^2 - b^2 = (a-b) (a+b)$

$(a-b)^2 = a^2 - 2ab + b^2$
Title: Re: quick 2 questions on simulanteous equations
Post by: Hielly on February 10, 2009, 04:34:56 pm: okay guys this is the same one as before but trying to find y.
equations are
ax+by=p
bx-ay=q

atm im at q(aq-bp)(p+b)/ap(-a^2-b^2)-b^3q
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 10, 2009, 04:37:37 pm: $ax+by=p$

$bx-ay=q$

times first one by b and second by a

$abx+b^2y=bp$

$abx-a^2y=aq$

take away

$b^2y+a^2y=bp-aq$

$y(a^2+b^2)= bp-aq$

$\implies y = \frac{bp-aq}{a^2+b^2}$
Title: Re: quick 2 questions on simulanteous equations
Post by: Hielly on February 10, 2009, 04:42:56 pm: omg flaming arrow, your a genius. It took me forever! Because when i found out x i just subbed it straight in equation 1, and then it gave me alot of pronumerals, that's what confused me. But now i know what to do, get rid of the x and then find y directly.
Title: Re: quick 2 questions on simulanteous equations
Post by: Flaming_Arrow on February 10, 2009, 05:10:19 pm: Quote from: Hielly on February 10, 2009, 04:42:56 pm
omg flaming arrow, your a genius. It took me forever! Because when i found out x i just subbed it straight in equation 1, and then it gave me alot of pronumerals, that's what confused me. But now i know what to do, get rid of the x and then find y directly.

yep thats the way.
Title: Re: yr 11 questions
Post by: Hielly on February 24, 2009, 06:36:46 pm: hey need help on this one

the amount of bending, Bmm, of a particular wooden beam under a load is given by B=0.2m^2 + 0.5m +2.5, where m kg is the mass (or load) on the end of the beam. what is the mass will produce a bend 8.8mm.

my equation is 8.8=0.2m^2 +0.5m +2.5
=0.2m^2 +0.5m-6.3

dont know what to do from here ( should i times them with 10 and then use completing the square?)
Title: Re: yr 11 questions
Post by: Flaming_Arrow on February 24, 2009, 07:28:54 pm: a= 2 b=5 c=-63

$\frac{-5 \pm \sqrt{25-4*2*-63}}{4}$

$\frac{-5 \pm \sqrt{529}}{4}$

$\frac{-5 \pm 23}{4}$

$x = \frac{9}{2} \ \mbox{or} \ x = -7$